The equation of motion of a particle are given by 

$\frac{dx}{dt}=t(t+1),\frac{dy}{dt}=\frac{1}{t+1}$

where the particle is at (x(t)), y(t) at time t. If the particle is at the origin at t = 0 then

$\frac{dx}{dt}=t(t+1)\Rightarrow x=\frac{t^3}{3}+\frac{t^2}{2}+C_1$

But at $t=0,x=0$ so $C_1=0$

Thus $x=\frac{t^3}{3}+\frac{t^2}{2}$                              $\left(1\right)$

$\frac{dy}{dt}=\frac{1}{t+1}\Rightarrow y=\log (t+1)+C_2$

But at $t=0,y=0$ so $C_2=0$ 

Thus $y=\log (t+1)\Rightarrow t=e^v-1$ 

Putting in (1), we have 

$\begin{array}{l}6x=t^2(2t+3)\\ ={\left(e^y-1\right)}^2\left(2e^y+1\right)\end{array}$